We will be using the little$-$endian convention that Qiskit adopts. This means that the left$-$most bit in any pair is the least significant. It also means that, when we draw a quantum circuit, the left$-$most bit will be the top line. Unless otherwise specified, qubits always start in the $|0$:$|$ state. Thus the circuit below is in the state $|01$:$|$ after the $x-$gate $($remember the $x-$gate switches the first qubit from $|0$:$|$ to $|1$:$|$

$90-x-$

$q_1$

Use tensor product multiplication and the identity matrix and the NOT $($or X$-$gate$)$ matrix to calculate the matrix representation of the circuit above. If the qubits start in the state $|00$:$|,$ use the matrix to show the state vector gives you the result you'd expect.